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May 4 Integrating Polar and Parametric Functions

Parametric Functions

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If you had this set of equations: $x = \cos{t}$, and $y = \sin{t}$, and you want to find the area of that ste of parametric equations. We could do
$$\text{The area is} \int\mathrm{y}\, \mathrm{d}x$$
$$x = \cos{t}, dx = -\sin{t}dt \rightarrow \int\mathrm{\sin{t}*-\sin{t}}\, \mathrm{d}t = \int\mathrm{-\sin^2{t}}\, \mathrm{d}t$$
$$\cos{2t} = 1 - 2\sin^2{t} \rightarrow -\sin^2{t} = \frac{1}{2}(cos{2t} - 1)$$
$$=\frac{1}{2}\int\mathrm{\cos{2t} - 1}\, \mathrm{d}t = \frac{\sin{2t}}{4} - \frac{x}{2} + C$$
$$\text{If you graph the system, you get that it is a unit circle. So, you can integrate from 0 to 2}\pi$$
$$(\frac{\sin{2t}}{4} - \frac{x}{2})|^{2\pi}_{0} = -\pi$$
$$\text{Note that that is the integral, not the area. The area is the absolute value, which is } \pi$$

Polar Equations

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You can think about doing area under a polar function as the sum of many sectors. The area of a sector is $\frac{1}{2\pi}(\pi{}r^2) = \frac{r^2}{2}$, so the area of the entire part is $\frac{1}{2}\int_{a}^{b}\mathrm{r(\theta{})^2}\, \mathrm{d}\theta{}$